Sum of squares

Number Theory Level 3

$Sum\quad of\quad first\quad n\quad terms\quad of\quad an\quad AP\quad is\quad c{ n }^{ 2 },\quad$ $then\quad the\quad sum\quad of\quad squares\quad of\quad these\quad n\quad terms\quad is$

\cfrac { n(4{ n }^{ 2 }+1){ c }^{ 2 } }{ 3 }

\cfrac { n(4{ n }^{ 2 }+1){ c }^{ 2 } }{ 6 }

\cfrac { n(4{ n }^{ 2 }-1){ c }^{ 2 } }{ 3 }

\cfrac { n(4{ n }^{ 2 }-1){ c }^{ 2 } }{ 6 }

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Sum of squares

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